How to calculate the volume of a skip bin container knowing the height of the material inside I need to know hot to calculate the volume of a skip bin (also known as a skip container or dumpster in some areas) with varying length and width.
It seems like a isosceles trapezoid when you look at it from the sides or the front. Like this:

The width of the skip varies only slightly, but the length below and above vary by about a yard.
Or like this:

The point is that in my work we put some semi-solids on it, and we use some aproximation to calculate the volume.
I'd like to know how to calculate the formula for exact value of the volume of liquid (i.e.) inside, but I keep thinking that it has something to do with integrals, infinitesimal sum, or something like it that exceeds my knowledge.
Anyone cares to help me, please?
 A: You certainly could use integration, but that would be tediously unnecessary. 
Thankfully, some simple geometry should suffice once you have some measurements:


*

*Find the area of the trapezoid face (left side on your picture) through the following formula:







*Then multiply that area by the "width" of the bin to find the total volume.


Conceptually, you could think of it as stacking up a bunch of those trapezoidal areas. 

The picture of the formula is from a Google search for "area of trapezoid"
A: The skip can be treated as a frustum of a rectangular right pyramid.
Let's say that the open face (larger) has dimensions $a \times b$, the base face (smaller) has dimensions $c \times d$, the skip has a vertical height between the opening and the base of $x$, and it's filled to height $h$ as measured from the base face with some material.
Then, the volume of the entire skip is:
$$V = \frac{xd}{3c}(a^2 + ac + c^2)$$
and the volume of the material is:
$$V_{fill} = \frac{cd}{3}\cdot [(\kappa^2 - 1)\frac{xc}{a-c} + \kappa^2h] $$
where $\displaystyle \kappa = \frac{ha - hc + xc}{xc}$
I'm almost certain the latter formula for the fill volume can be further simplified, but I'll leave that to you.
A: Let:

$l$ – length of the base
  $w$ – width of the base
  $\Delta l$ – diference between the lengths of the top and the base
  $\Delta w$ – diference between the widths of the top and the base
  $H$ – height of the figure

Then the volume:
$$V=\left(lw + \frac{l\Delta w + w\Delta l}2 + \frac {\Delta l\Delta w}3\right)H$$
